(A) (I) Line in Correct Position from T = 0 to T =14 Min; 1

(a) (i) line in correct position from t = 0 to t =14 min; 1

Allow ± square.

(ii) reasonable tangent drawn at correct position ie at 76°C but allow
70°C ® 85°C;

If tangent position is not acceptable, award [1 max] in this section
for length of tangent.

line length used for tangent at least 8 cm;

value 0.09 ® 0.13;

accuracy - value within 0.10 ® 0.12; 4

Award [1] for value between ± 10% and ± 20% of 0.11 and [2] for
value 0.11 ± 10%.

Unit of answer is not required. If the candidate fails to convert to
s-1 , then award one of the last two marks for answer in range
6.0 ® 7.2°C min-1 .

(b)

(i) point plotted correctly; (allow ecf from (b)) 1

(ii) error bar at qE = 20°C: 4(± 2)mm long;

error bar at qE = 81°C : 20(± 4)mm long; 2

Ignore any horizontal error bars.

points lie on straight-line;

and line passes through origin; 3

Award [1] for “last point off line, so not obeyed”.

[11]

2. Linear motion

(a) (i)

= 1.9 ´ 104 J; 2

(ii) uses area between the t-axis and the line;

correctly converts area ® distance (one 1cm ´ 1cm square º 5.0m);

distance between 90 m and 105 m;

improved accuracy, distance between 95 m and 100 m; 4

Do not accept kinematic formulas. Distance can only be found
from area.

(b) (i) DEP = 72 ´ 9.8 ´ 41;

= 2.9 ´ 104 J; 2

Accept 3.0 ´ 104J for responses using g = 10ms-2.

(ii) energy “loss” = 1.0 ´ 104 J;

average force =

= 100 N; 3

NB follow through working - answer is {(b)(i)–(a)(i)} / (a)(ii).

(iii) eg air resistance;

friction between skis and slope;

force to push snow away from skis; 2

To award marks responses must specify where friction is acting.

time of flight = 0.61s;

horizontal distance travelled(= 23 ´ 0.61) = 14m;

distance CD(= 14 - 12)= 2.0m; 4

Accept a time of 0.60 s and CD = 1.8 m for responses using g = 10ms-2.

(d) (i) D is further from the edge C; 1

(ii) sensible reason eg velocity not normal to ground;

hence impact is less; (any other sensible comment) 2

[20]

3. (a) more energetic molecules leave surface;
mean kinetic energy of molecules in liquid decreases;
and mean kinetic energy depends on temperature; 3

Award [2] if mean not mentioned.

(b) eg larger surface area;
increased draught;
higher temperature;
lower vapour pressure; 2 max

Award [1] if candidate merely identifies two factors.

Allow ecf if one term incorrect or missing.

[9]

4. (a) (i) distance travelled per unit time;
by the energy of the wave / by a wavefront; 2

(ii) velocity has direction; but light travels in all directions; 2

(b) (i) distance in a particular direction; (accept in terms of energy transfer)
(of a particle) from its mean position; 2

(ii) longitudinal: displacement along;
transverse: displacement normal to;
direction of transfer of wave energy / propagation, not motion; 3
Award [0] for left / right and up / down for longitudinal / transverse.

(ii) = 5.8 km s–1; (±0.1) 1

Award [1 max] if the answers to (i) and (ii) are given in reversed order.

(d) (i) P shown as the earlier (left hand) pulse; 1

(ii) laboratory L3; 1

(iii) eg pulses arrive sooner;
smaller S-P interval;
larger amplitude of pulses; 3

Allow any feasible piece of evidence, award [1] for each up to [3 max].

(iv) distance from L1 = 1060 km; (± 20)
distance from L2 = 650 km; (± 20)
distance from L3 = 420 km; (± 20)

Accept 3 significant digits in all three estimates.

some explanation of working; 4

(v) position marked, consistent with answers to (iv);
to the right of line L2L3, closer to L3; 1 max

If the answers given in (iv) means that the point cannot be plotted, then only allow the mark if the candidate states that the position cannot be plotted / does not make sense.

(e) (i) illustration showing node at centre, antinode at each end; 1

(ii) wavelength of standing wave = (2 × 280) = 560 m / ecf
or = 570 m;
frequency = » 6Hz
or wavelength of standing wave = (2 × 280) = 560 m;
earthquake frequency is natural frequency of vibration of
building / mention of resonance / multiple / submultiple if ecf; 3

[25]

5. The properties of sound waves

(a) substitution into speed = distance / time to get

distance =1500 ´ 0.012 = 18m;

therefore, depth = 18 ¸ 2 = 9m; 2

(b) (i) appropriate wavefronts shown in geometric shadow region;

of constant wavelength; 2

(ii) substitution of correct values into v = fl

to get l = 0.025m;

so obstacle (fish) significantly larger than wavelength hence
diffraction effect small / OWTTE; 2

or the wavelength is very small;

so diffraction will be small for any reasonably sized fish;

as a result of relative motion of source and observer; 2

Accept other general descriptions but award [1 max] for an answer
that just gives an example of the Doppler effect.

(ii)

to get v = 30ms-1;

or:

with f = 450Hz;

justification of f = 450Hz;

to get v = 28ms-1;

or:

with f = 450Hz;

justification of f = 450Hz;

to get v = 33ms-1; 4

[12]

6. (a) shape of diffraction pattern acceptable;

central maximum of one pattern falls on first minimum of other;

relative heights of central and first maxima realistic for both patterns; 3

(b)

woman ® car distance = 3

= 7.4 km

[6]